 Spherical turnaround from an unperturbed Hubble velocity

In an Einstein-de Sitter universe, give an initially spherical, pressureless region total mass $M$ and excess mass $0<\delta M\ll M$, while holding its initial expansion velocity at the unperturbed Hubble flow value at physical radius $r_0$. Before shell crossing, the outer shell has ${\dot{r}}_0^2=2G(M-\delta M)/r_0$ and conserved specific orbital energy $-G\delta M/r_0$. The shell theorem and conservation of energy give ${\dot{r}}^2=2GM/r-2G\delta M/r_0$, so its cosmological turnaround radius is $M{r}_0/\delta M$. With $A=2GM$, $B=2G\delta M/r_0$ and $ϵ=\delta M/M$, the elapsed cosmic time is exactlyReplacing the square bracket by $\pi /2$ gives the displayed approximation, with relative overestimate $4{ϵ}^{3/2}/(3\pi )+O({ϵ}^{5/2})$. Holding the initial expansion velocity fixed is an essential condition: a pure growing-mode linear cosmological density perturbation has an initial peculiar velocity as well as an excess mass, changing the binding energy at the same order.